find the value of k so that the straight line 2x+3y+4+k(6x-y+12)=0 is perpendicular to 7x+5y-4=0
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2x+3y+4+k(6x-y+12) = 0
2x+3y+4+6kx-ky+12k = 0
2x+6kx+3y-ky+12k+4 = 0
2x(1+3k)+y(3-k)+4(3k+1) = 0
y(3-k) = -2x(1+3k)-4(3k+1)
y = -2x(1+3k)/(3-k)-4(3k+1)
Therefore, slope is -2(1+3k)/(3-k)
7x+5y-4 = 0
5y = -7x+4
Therefore, slope is -7/5
If lines are perpendicular then their product is -1
So,
2(1+3k)/(3-k)x(-7/5) = -1
2(1+3k) = 5(3-k)/7
1+3k = 15-5k/14
14+42k = 15-5k
42k+5k = 15+14
47k = 29
k = 29/47
2x+3y+4+6kx-ky+12k = 0
2x+6kx+3y-ky+12k+4 = 0
2x(1+3k)+y(3-k)+4(3k+1) = 0
y(3-k) = -2x(1+3k)-4(3k+1)
y = -2x(1+3k)/(3-k)-4(3k+1)
Therefore, slope is -2(1+3k)/(3-k)
7x+5y-4 = 0
5y = -7x+4
Therefore, slope is -7/5
If lines are perpendicular then their product is -1
So,
2(1+3k)/(3-k)x(-7/5) = -1
2(1+3k) = 5(3-k)/7
1+3k = 15-5k/14
14+42k = 15-5k
42k+5k = 15+14
47k = 29
k = 29/47
afhjafeef4:
k=-29/37 m1=-2-6k/3-k m2=-7/5
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